20cm(cube) of a gaseous hydrocarbon were mixed with 90cm(cube) oxygen, and the mixture exploded. At room temperature, 60cm(cube) of gas were left. 40cm(cube) of the gas(cabon(4)oxide) were absorbed by sodium hydroxide, leaving 20cm(cube) of oxygen. find the empirical formular of the hydrocarbon.
Write some of the hydrocarbon rxns with oxygen. All give CO2 + H2O. The problem would be easier if you were given cc H2O (as a gas) but if you don't have that I would do this.
CH4 + 2O2 ==> C2 + 2H2O
C2H6 + 7O2 ==> 4CO2 + 6H2O
C3H8 + 5O2 --> 3CO2 + 4H2O
2C4H10 + 13O2 ==> 8CO2 + 10H2O
etc. Now what did you have?
CxHy + O2 ==> xCO2 + (y/2)H2O
20 cc CxHy
O2 used must be 90-20 = 70 cc.
CO2 produced = 40 cc
The ratio of these numbers is
CxHy + O2 ....> xCO2 + (y/2)H2O
..20...70.......40........? OR
1.....3.5.......2.......?
That means x must 2 but what is H2O. If you have it in the problem that makes it easier. If you don't have it then you show that of all of the combustions I wrote at the beginning the one with C2H6 is the only one that fits. That is the ONLY one with CO2 twice the hydrocarbon and reactions I didn't write for saturated hydrocarbons become more divergent. Note that H2O in C2H6 (when combusted) is 3x that of the hydrocarbon. So y/2 = 3 and y = 6 and CxHy is C2H6.
Well, let's break it down, or should I say, "ignite" our investigation! We have 20 cm³ of the mysterious hydrocarbon and 90 cm³ of oxygen. After the explosion, we're left with 60 cm³ of gas.
Now, it's time to play a little game of process of elimination. We know that 40 cm³ of the gas is absorbed by sodium hydroxide, leaving only 20 cm³ of oxygen behind. So, we can confidently say that the remaining 20 cm³ of gas must be the hydrocarbon.
Now, to find the empirical formula, we need to figure out the ratio of carbon to hydrogen in the hydrocarbon. Since we don't have any information about the carbon to hydrogen ratio, we'll have to use some wizardry to make an educated guess.
Let's assume the simplest ratio of carbon to hydrogen, which is 1:2. This means we have one carbon atom and two hydrogen atoms in our hydrocarbon.
Given that the final 20 cm³ of gas is the hydrocarbon, we can compare it to the 60 cm³ of remaining gas to determine the ratio. If we take 20 cm³ out of the 60 cm³ of gas, we're left with 40 cm³ of oxygen.
This means that the ratio of the hydrocarbon to oxygen is 20:40, which simplifies to 1:2.
So, based on our assumption of a 1:2 carbon to hydrogen ratio and the empirical formula of the hydrocarbon being C₁H₂, we can conclude that the empirical formula of the hydrocarbon is... drumroll please... C₁H₂!
To find the empirical formula of the hydrocarbon, we need to determine the mole ratios of the elements present in the compound.
Let's start by converting the given volumes of gases to the number of moles of each gas using the ideal gas law.
1. Calculate the number of moles of oxygen (O2) initially present:
V = nRT, where V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature (assumed constant at room temperature)
n(O2) = (90 cm^3) / (22.4 cm^3/mol) (using the ideal gas volume at STP condition)
= 4 moles (approximately)
2. Calculate the number of moles of oxygen remaining:
n(O2 remaining) = (20 cm^3) / (22.4 cm^3/mol)
= 0.89 moles (approximately)
3. Calculate the number of moles of carbon dioxide (CO2) formed:
n(CO2) = (40 cm^3) / (22.4 cm^3/mol)
= 1.79 moles (approximately)
4. Calculate the number of moles of carbon, assuming it came from the hydrocarbon:
n(C) = n(CO2) (since 1 mole of CO2 contains 1 mole of C)
= 1.79 moles
5. Calculate the number of moles of hydrogen, assuming it came from the hydrocarbon:
n(H) = [n(O2) - n(O2 remaining)] / 2 (since 1 mole of O2 reacts with 2 moles of H)
= [4 - 0.89] / 2
= 1.56 moles (approximately)
6. Find the mole ratio of carbon to hydrogen:
Divide the number of moles of carbon and hydrogen by the smaller value (1.56 moles in this case):
C: 1.79 moles / 1.56 moles = 1.15 (approximately)
H: 1.56 moles / 1.56 moles = 1.00
Based on the resulting mole ratios, we can round the values to the nearest whole number to determine the empirical formula of the hydrocarbon. In this case, the empirical formula would be CH.
To find the empirical formula of the hydrocarbon, we need to analyze the given information step by step.
Step 1: Determine the volume of the remaining gas after the explosion.
We start with 20 cm³ of the gaseous hydrocarbon and 90 cm³ of oxygen. So, the initial volume of the mixture is 20 cm³ + 90 cm³ = 110 cm³.
After the explosion, we are left with 60 cm³ of gas. This indicates that 110 cm³ - 60 cm³ = 50 cm³ of gas has been consumed.
Step 2: Determine the volume of carbon dioxide produced.
The remaining 50 cm³ of gas consists of the product after the combustion. By deducting the volume of oxygen left (20 cm³) from the remaining gas (50 cm³), we find that 50 cm³ - 20 cm³ = 30 cm³ of carbon dioxide (CO₂) has been produced.
Step 3: Determine the volume of oxygen consumed during combustion.
Since the reaction produces carbon dioxide (CO₂) from the hydrocarbon, the volume of the consumed oxygen is equal to the volume of carbon dioxide produced. Therefore, the volume of oxygen consumed is also 30 cm³.
Step 4: Determine the volume of hydrogen produced.
The remaining gas after the combustion is primarily hydrogen (H₂). By subtracting the volume of carbon dioxide produced (30 cm³) and the volume of oxygen left (20 cm³) from the remaining gas (60 cm³), we find that 60 cm³ - 30 cm³ - 20 cm³ = 10 cm³ of hydrogen has been produced.
Step 5: Determine the moles of carbon, hydrogen, and oxygen.
To calculate the empirical formula, we need to find the moles of carbon, hydrogen, and oxygen. Using the ideal gas law, we can relate the volume of a gas to its number of moles.
We know that 1 mole of gas occupies 22.4 dm³ at standard temperature and pressure (STP), which is equivalent to 22,400 cm³.
Using this information, we can calculate the moles of each element present:
- Moles of carbon (C): (30 cm³ / 22,400 cm³/mol) = 0.00134 mol
- Moles of hydrogen (H): (10 cm³ / 22,400 cm³/mol) = 0.00045 mol
- Moles of oxygen (O): (20 cm³ / 22,400 cm³/mol) = 0.00089 mol
Step 6: Determine the empirical formula.
Now that we have the number of moles of each element, we need to simplify the ratio to obtain the simplest whole number ratio for the empirical formula.
Dividing all the mole values by the smallest value (0.00045 mol) results in the following ratio:
C ≈ 0.00134 mol / 0.00045 mol ≈ 2.98
H ≈ 0.00045 mol / 0.00045 mol ≈ 1
O ≈ 0.00089 mol / 0.00045 mol ≈ 1.98
Rounding these values to the nearest whole number, we get the following ratio:
C2H2O
Therefore, the empirical formula of the hydrocarbon is C2H2O.